Why is $\sigma\tau\neq\tau\sigma$ in $\operatorname{Gal}(\mathbb{Q}(\sqrt[3] 2,\zeta_3)/\mathbb{Q})?$

Question

Theorem: $\operatorname{Gal}(\mathbb{Q}(\sqrt[3] 2,\zeta_3)/\mathbb{Q}) =\langle\sigma, \tau\rangle\cong S_3$

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My question: Why is $\sigma\tau\neq\tau\sigma$ in $\operatorname{Gal}(\mathbb{Q}(\sqrt[3] 2,\zeta_3)/\mathbb{Q})?$

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My attempt: I take $\sigma(\sqrt[3] 2, \zeta_3)=( \sqrt[3]2\zeta_3, \zeta_3)$ and $\tau(\sqrt[3] 2, \zeta_3)=(\sqrt[3] 2,\zeta_3^2)$

Then $\sigma\tau(\sqrt[3] 2, \zeta_3)=\sigma (\sqrt[3]2,\zeta_3^2)=( \sqrt[3]2\zeta_3, \zeta_3^2)\tag1$

similarly $ \tau \sigma(\sqrt[3] 2, \zeta_3) =\tau( \sqrt[3]2\zeta_3, \zeta_3)=( \sqrt[3]2\zeta_3, \zeta_3^2)\tag2$

From $(1)$ and $(2) $

we have $$\sigma\tau=\tau\sigma$$

This implies $\operatorname{Gal}(\mathbb{Q}(\sqrt[3] 2,\zeta_3)/\mathbb{Q}$ is not isomorphic to $S_3$

Tell me where I'm wrong?

Answer 1

Well,

$$ \sigma\tau(\sqrt[3]2,\zeta_3)=\sigma(\sqrt[3]2,\zeta_3^2)=(\sqrt[3]2\zeta_3^\color{red}1,\zeta_3^2) $$

while

$$ \tau\sigma(\sqrt[3]2,\zeta_3)=\tau(\sqrt[3]2\zeta_3,\zeta_3)=(\sqrt[3]2\zeta_3^\color{red}2,\zeta_3^2)\,. $$